I found them in a test but just able to solve and there are no answers announced, so please check my answer for
$$I(2,n)\ ,\ I(3,n)\ ,\ I(5,3)$$ given $$I(m,1) =I(1,n) =1\ and\ I (m+1,n) +I (m,n+1) =I (m+1,n+1)$$
These are my answer $$I(2,n) = {n}$$ $$I(3,n) = \frac{n(n+1)}{2}+1$$ $$I(5,3) = 15$$
$$I(m,n) = \binom {n+m-2}{n-1} = \frac {(n+m-2)!}{(n-1)!(m-1)!},$$ so the first and the third answers are correct while the second is not. $I(3,n)=\frac {n(n+1)} 2$.